Given:
n = 12 moles of oxygen
T = 273 K, temperature
p = 75 kPa, pressure
Use the ideal gas law, given by
[tex]pV=nRT \\ or \\ V= \frac{nRT}{p} [/tex]
where
V = volume
R = 8.3145 J/(mol-K), the gas constant
Therefore,
[tex]V= \frac{(12\,mol)(8.3145\, \frac{J}{mol-K} )(273\,K)}{75 \times 10^{3} \, Pa}= 0.3632\,m^{3}[/tex]
Answer: 0.363 m³
